Originally Posted by

**michaela-donnelly** Hi all

I have a problem that I've nearly finished but I'm stuck on the last bit so I'm hoping someone could help out. The question is

$\displaystyle \int\int (y^2 z^2 + z^2 x^2 + x^2 y^2) \: dS$

Evaluated on the region of $\displaystyle z^2 = x^2 + y^2$ between z=1 and z=2.

My workings so far are....

With z=√(x²+y²), the integrand should become

$\displaystyle y^2 (x^2 + y^2) + x^2 (x^2 + y^2) + x^2 y^2$

which simplifies to $\displaystyle (x^2 + y^2)^2 + x^2 y^2$

Now I have

$\displaystyle \sqrt{2} \int \int ( (x^2 + y^2)^2 + x^2 y^2) \: dA$

To evaluate using polar coordinates, I need to determine the limits of integration.

Since z ranges from 1 to 2 and the equation of the cone is z=r, I have $\displaystyle 1 \leq r \leq 2$, and $\displaystyle 0\leq \theta \leq 2\pi$

By substituting in x=r*cos(θ) and y=r*sin(θ), and multiplying by 'r' (the Jacobian determinant), I get

$\displaystyle \sqrt{2} \int_{0}^{2\pi} \int_{1}^{2}(r^5 + r^3 \cos(\theta)^2 \sin(\theta)^2) \: dr \: d\theta$

i think the best way is to integrate WRT r first:

$\displaystyle \sqrt{2} \int_{0}^{2\pi} \int_{1}^{2}(r^5 + r^3 \cos(\theta)^2 \sin(\theta)^2) \: dr \: d\theta$

(im going to leave out the $\displaystyle \sqrt{2}$ for now)

$\displaystyle \Rightarrow \int_{0}^{2\pi} [\frac{r^6}{6} + \frac{r^4}{4}]_{1}^{2} \cos(\theta)^2 \sin(\theta)^2) \: d\theta$

$\displaystyle \Rightarrow \int_{0}^{2\pi} [\frac{32}{3} + 4 \cos(\theta)^2 \sin(\theta)^2)] - [\frac{1}{6} + \frac{1}{4} \cos(\theta)^2 \sin(\theta)^2)] \: d\theta$

I know I then simplify this and integrate WRT theta but this is where I'm running into difficulties so any help at all would be so great.

Thanks